Solution to puzzle 81: Digit transfer

Find the smallest positive integer such that when its last digit is moved to the start of the number (example: 1234 becomes 4123) the resulting number is larger than and is an integral multiple of the original number. Numbers are written in standard decimal notation, with no leading zeroes.

Suppose the n-digit integer s = a₁a₂a₃...a_n is multiplied by k when the digit a_n is transferred to the beginning of the number. (That is, t = a_na₁a₂...a_n−1 = ks, where t is the resulting number.)
Note that we must have a₁ > 0, since s is written with no leading zeroes; and a_n > 1, so that when a_n is transferred to the beginning of the number, the resulting number is two or more times the original number.

Consider the infinite repeating decimals x = 0.a₁a₂a₃...a_na₁a₂a₃...a_n... and y = 0.a_na₁a₂...a_n−1a_na₁a₂...a_n−1... , formed by repeating s and t, respectively.

We have 0.a₁a₂a₃...a_n = s/10ⁿ, and so x = s/(10ⁿ − 1). (This follows by considering x as the sum to infinity of a geometric series.)
Similarly, we have 0.a_na₁a₂...a_n−1 = t/10ⁿ = ks/10ⁿ, and so y = ks/(10ⁿ − 1).

Hence y = kx.

Clearly, we also have y = a_n/10 + x/10, or 10y = a_n + x.
Therefore 10kx = a_n + x, from which x = a_n/(10k − 1).

We have thus reduced the problem to one of trial and error for various values of a_n and k, neither of which can be greater than 9.
We must also have a_n k. For each value of k, we obtain the smallest value of x (and therefore of s) when a_n = k.

Testing each value of k/(10k − 1), for 2 k 9, we find the smallest s occurs for k = 4, when 4/39 yields 102564.

Hence the smallest positive integer such that when its last digit is moved to the start of the number the resulting number is larger than and is an integral multiple of the original number, is 102564.

Source: Traditional